August 15th, 2018, 06:32 PM  #1 
Senior Member Joined: Jan 2012 Posts: 137 Thanks: 2  Evaluating a limit
Hi, Please evaluate the following limit: $\displaystyle \lim_{x>0}\left ( \tfrac{1}{x^{2}}\cot^{2}x \right )$ Thanks. Last edited by skipjack; August 15th, 2018 at 07:41 PM. 
August 15th, 2018, 07:03 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 621 Thanks: 394 Math Focus: Dynamical systems, analytic function theory, numerics 
The limit is 2/3.

August 15th, 2018, 07:27 PM  #3 
Senior Member Joined: Jan 2012 Posts: 137 Thanks: 2  Last edited by skipjack; August 15th, 2018 at 08:29 PM. 
August 15th, 2018, 07:44 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra 
Assuming that the limit exists and that therefore $$\lim_{x \to 0} \left(\frac1{x^2}\cot^2{x}\right) = L$$ setting $2t=x$ we have that \begin{align}L &= \lim_{t \to 0} \left(\frac1{(2t)^2}\cot^2{2t}\right) \\ &= \lim_{t \to 0} \left(\frac1{4t^2}\frac{\cos^2{2t}}{\sin^2{2t}}\right) \\ &= \lim_{t \to 0} \left(\frac1{4t^2}\frac{\cos^2{2t}}{(2\sin{t}\cos{t})^2}\right) \\ &= \lim_{t \to 0} \left(\frac1{4t^2}\frac{(\cos^2{t}\sin^2{t})^2}{4\sin^2{t}\cos^2{t}}\right) \\ 4L &= \lim_{t \to 0} \left(\frac1{t^2}\frac{\cos^4{t}2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\ 3L &= \lim_{t \to 0} \left(\frac1{t^2}\frac{\cos^4{t}2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right)  \lim_{t \to 0} \left(\frac1{t^2}\cot^2{t}\right) \\ &=\lim_{t \to 0} \left(\frac1{t^2}\frac{\cos^4{t}2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}}  \frac1{t^2}+\cot^2{t}\right) \\ &=\lim_{t \to 0} \left(\frac{\cos^2{t}}{\sin^2{t}}  \frac{\cos^4{t}2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\ &= \lim_{t \to 0} \left(\frac{\cos^4{t}}{\sin^2{t}\cos^2{t}}  \frac{\cos^4{t}2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\ &= \lim_{t \to 0} \left(\frac{2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{ t}\cos^2{t}} \right) \\ &= \lim_{t \to 0} \left(2+\tan^2{t} \right) \\ &=2 \\ L&=\frac23 \end{align} Thanks to this for inspiration. 
August 15th, 2018, 08:21 PM  #5  
Senior Member Joined: Jan 2012 Posts: 137 Thanks: 2  Quote:
Thx.  
August 15th, 2018, 09:06 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,747 Thanks: 2133 
$\displaystyle \begin{align*}\cot^2\!x &= 1/(\sin^2\!x)  1 \\ &= 1/(x^2(1  x^2/3 \,+\; ...))  1 \\ &= (1/x^2)(1 + x^2/3 \,+\; ...)  1 \\ &= 1/x^2  2/3 \,+\; ...\end{align*}$ so the required limit is 2/3. 
August 16th, 2018, 05:06 AM  #7 
Senior Member Joined: Sep 2016 From: USA Posts: 621 Thanks: 394 Math Focus: Dynamical systems, analytic function theory, numerics  
August 16th, 2018, 09:59 AM  #8 
Senior Member Joined: Jan 2012 Posts: 137 Thanks: 2  

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